6 Answers

Probability of at least one means union of the probability of events, i.e.,
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
here, $P(A\cup B) = 1$ , because it can not be more than $1$ and if at least one of the event has probability $1$ (here , $P(A) = 1$), then union of both should be $1.$
So,
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$

Probability of A given that B has happened, P(A|B)= 1
it has to be one. BECAUSE. Probability of A is 1
and It will still be 1 even if B has happened(unless B reduces it)
Since A and B(intersection) is non zero.

First of all the divison must be half because p(A/B) / p(B/A)=1 . 1/2 so the answer must be option c or option d ..now p(A/B) is always 1 as the p(A)=1 and it doesnt depend any other event so option d is correct

Say we are living in an area where it rains everyday (Event A) and there are two groups of day MWF and TTS (Just assume that Sunday doesn't exist at all). Let Event B be that the chosen day is one amongst MWF. Clearly the event A has probability of 1 and probability of event B is 1/2.

Now the question can be interpreted as :

P(A|B) = Given that the day chosen is among MWF, probability of raining ?

Since it is raining no matter what, therefore the probability of this part is 1.

P(B|A) = Given that it is raining, probability that the day is one of MWF ?

Since it rains on all days, the probability that the randomly chosen day is one amongst MWF is 3/6 or 1/2.